The Jordan-Hölder Theorem

The Jordan-Hölder Theorem

We now define the following.

Definition 1 An \(A\)-module \(M\) is called simple if \(M\neq 0\) and the only submodules of \(M\) are \(0\) and \(M\).

It is obvious that a simple module must be generated by a single element \(x\in M\), but considering \(\mathbb{Z}/6\mathbb{Z}\), the converse does not hold. On the other hand, if a simple module \(M\) is generated by \(x\), then defining the \(A\)-module homomorphism \(A \rightarrow M\) by \(1\mapsto x\), the isomorphism

\[A/\ann(M)=A/\ann(x)\cong M\]

is obtained. If \(\ann(M)\) is not a maximal ideal, then with \(\ann(M)\) contained in a maximal ideal of \(A\), \(\mathfrak{m}\), \(\mathfrak{m}/\ann(M)\) would be a submodule of \(M\), so it is also obvious that \(\ann(M)\) must be a maximal ideal of \(A\).

Definition 2 Fix an \(A\)-module \(M\). A decreasing sequence of submodules of \(M\)

\[M=M_0\supsetneq M_1\supsetneq \cdots\supsetneq M_n=0\]

is called a chain of length \(n\). This chain is called a composition series if for every \(k\), \(M_k/M_{k+1}\) is a simple module. The smallest length among such composition series is called the length of \(M\), denoted \(\length(M)\). If no composition series of \(M\) exists, we define \(\length(M)=\infty\).

Then the following theorem can be regarded as the module version of the Jordan-Hölder theorem.

Theorem 3 An \(A\)-module \(M\) has a finite composition series if and only if \(M\) is both Artinian and Noetherian. Assume that this condition is satisfied and that a composition series of length \(n\)

\[M=M_0\supsetneq M_1\supsetneq \cdots\supsetneq M_n=0\]

is given. Then the following hold.

1. Any chain of length at most \(n\) of submodules of \(M\) admits a refinement to a composition series.
2. For the collection of maximal ideals such that \(M_k/M_{k+1}\cong A/\mathfrak{m}\) for some \(k\), there exists an isomorphism \(M\cong\bigoplus_{\mathfrak{m}}M_\mathfrak{m}\).
3. If for some maximal ideal \(\mathfrak{p}\) and \(k\), \(\mathfrak{p}^k\) annihilates \(M\), then \(M=M_\mathfrak{p}\).

Proof

First, suppose that \(M\) is both Artinian and Noetherian. To begin with, from the condition that \(M\) is Noetherian, we can find for \(M\) a suitable maximal proper submodule \(M_1\). Meanwhile, since it is clear that a submodule of a Noetherian module must be Noetherian, repeating this allows us to find for \(M_k\) a maximal proper submodule \(M_{k+1}\). The chain thus defined,

\[M=M_0\supsetneq M_1\supsetneq \cdots\]

has finite length by the Artinian condition, and since \(M_{k+1}\) is a maximal proper submodule of \(M_k\), we know that this chain is a composition series.

The first result is proved in the same way as the Jordan-Hölder theorem, so we do not prove it separately. Now, once this is accepted, whenever an arbitrary chain is given, we can refine this chain to a composition series, and thus we can even show the reverse direction of the preceding equivalence.

We now prove the second result. From the finiteness of the given chain, we know that there are only finitely many maximal ideals satisfying the condition, so \(\bigoplus_\mathfrak{m} M_\mathfrak{m}\) can be viewed as \(\prod_\mathfrak{m} M_\mathfrak{m}\); here, the given map is obtained by applying the universal property of the direct product to the maps \(M \rightarrow M_\mathfrak{m}\). To show that this map is an isomorphism, it suffices to apply §Properties of Localization, ⁋Proposition 4 and examine the localization at a maximal ideal.

For this, first, if \(M\cong A/\mathfrak{m}\), then for any maximal ideal \(\mathfrak{m}'\),

\[M_{\mathfrak{m}'}=\begin{cases}M&\text{if $\mathfrak{m}=\mathfrak{m}'$,}\\0&\text{otherwise.}\end{cases}\]

From this, given a composition series of \(M\)

\[M=M_0\supsetneq M_1\supsetneq \cdots\supsetneq M_n=0\]

localizing at a maximal ideal \(\mathfrak{m}\) yields

\[M_\mathfrak{m}=(M_0)_\mathfrak{m}\supseteq (M_1)_\mathfrak{m}\supseteq \cdots\supseteq (M_n)_\mathfrak{m}=0\]

However, combining the fact that the localization functor is an exact functor (§Properties of Localization, ⁋Proposition 2) with the computation above,

\[(M_k)_\mathfrak{m}/(M_{k+1})_\mathfrak{m}\cong (M_k/M_{k+1})_\mathfrak{m}=\begin{cases}M_k/M_{k+1}&\text{if $M_k/M_{k+1}\cong A/\mathfrak{m}$,}\\0&\text{otherwise}\end{cases}\]

holds. From this we obtain the second result, and the third result can be proved similarly to the above computation.

Properties of Artinian and Noetherian Rings

Theorem 4 For a ring \(A\), the following are all equivalent.

1. \(A\) is Noetherian and every prime ideal is maximal.
2. \(A\) has finite length as an \(A\)-module.
3. \(A\) is Artinian.

Proof

First, assume the first condition and show the second condition. Suppose for contradiction that \(A\) satisfies the first condition but does not have finite length. Let \(\mathfrak{a}\) be maximal among the ideals of \(A\) such that \(A/\mathfrak{a}\) does not have finite length. Then if \(ab\in \mathfrak{a}\) and \(a\not\in \mathfrak{a}\), we can consider the following short exact sequence

\[0\longrightarrow A/(\mathfrak{a}:(a))\overset{a}{\longrightarrow}A/\mathfrak{a}\longrightarrow A/(\mathfrak{a}+(a))\longrightarrow 0\]

(§Basic Notions, §§Basic Definitions). Now, since \(\mathfrak{a}+(a)\) is an ideal strictly containing \(\mathfrak{a}\), by definition \(A/(\mathfrak{a}+(a))\) must have finite length. On the other hand, from the definition we have \(\mathfrak{a}\subseteq (\mathfrak{a}:(a))\), and if \(b\not\in \mathfrak{a}\), then \(\mathfrak{a}\subsetneq (\mathfrak{a}:(a))\), so likewise by the definition of \(\mathfrak{a}\), \(A/(\mathfrak{a}:(a))\) must have finite length. Therefore, through these composition series we obtain a composition series for \(A/\mathfrak{a}\), and since this contradicts the assumption that \(A/\mathfrak{a}\) does not have finite length, we have \(b\in \mathfrak{a}\). That is, \(\mathfrak{a}\) is a prime ideal. Now, by the assumption of condition 1, \(\mathfrak{a}\) is maximal, so \(A/\mathfrak{a}\) is a field, which again contradicts the assumption that \(A/\mathfrak{a}\) does not have finite length; thus we obtain the desired result.

Now, assuming the second condition, the third condition is immediate from Theorem 3. Therefore, it suffices to assume the third condition and show the first condition. To this end, consider the collection of ideals obtained as products of maximal ideals of \(A\). Then, since \(A\) is Artinian, there exists a minimal ideal \(\mathfrak{a}\) in this collection. Then \(\mathfrak{a}=0\), and therefore the zero ideal can be written as a product of maximal ideals \(0=\mathfrak{m}_1\cdots\mathfrak{m}_k\).

To prove this claim, first observe that the minimality of \(\mathfrak{a}\) means that \(\mathfrak{m}\mathfrak{a}=\mathfrak{a}\) for every maximal ideal \(\mathfrak{m}\). That is, \(\mathfrak{a}\subseteq \mathfrak{m}\). By similar logic, since \(\mathfrak{a}^2\) is a product of maximal ideals, \(\mathfrak{a}^2=\mathfrak{a}\) also holds. Now suppose for contradiction that \(\mathfrak{a}\neq 0\). Then there exists an ideal such that \(\mathfrak{a}\mathfrak{b}\neq 0\), and again from the assumption that \(A\) is Artinian, we can choose \(\mathfrak{b}\) to be minimal among the ideals satisfying this property. Then since

\[(\mathfrak{b}\mathfrak{a})\mathfrak{a}=\mathfrak{b}\mathfrak{a}^2=\mathfrak{b}\mathfrak{a}\neq 0\]

from the minimality of \(\mathfrak{b}\) we know that we must have \(\mathfrak{b}\mathfrak{a}=\mathfrak{b}\).

Now from the definition of \(\mathfrak{b}\), we know that \(y \mathfrak{a}\neq 0\) must hold for some \(y\in \mathfrak{b}\), and from the minimality of \(\mathfrak{b}\) we know that we must have \(\mathfrak{b}=(y)\). Now from the above equality \(\mathfrak{b}\mathfrak{a}=\mathfrak{b}\), we know that \(xy=y\) must hold for some \(x\in \mathfrak{a}\). That is, \((1-x)y=0\) must hold. But \(x\in \mathfrak{a}\), and since \(\mathfrak{a}\) is contained in every maximal ideal, \(1-x\) is not contained in any maximal ideal. That is, \(1-x\) is a unit, from which \(y=0\), and this contradicts the definition of \(\mathfrak{b}\). That is, \(\mathfrak{a}=0\), and therefore the zero ideal can be written as a product of maximal ideals \(0=\mathfrak{m}_1\cdots\mathfrak{m}_k\).

Formally setting \(\mathfrak{m}_1\cdots\mathfrak{m}_0=A\), for each \(l=0,\ldots, k-1\), if we view \(\mathfrak{m}_1\cdots\mathfrak{m}_l/\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\) as an \(A/\mathfrak{m}_{l+1}\)-vector space, a submodule of this vector space can be regarded as an ideal of \(A\) containing \(\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\) and contained in \(\mathfrak{m}_1\cdots\mathfrak{m}_l\), and since \(A\) is Artinian, from this we know that \(\mathfrak{m}_1\cdots\mathfrak{m}_l/\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\) is finite-dimensional. Now, collecting all of these together, we obtain a composition series for \(A\), and therefore by Theorem 3, \(A\) is Noetherian.

To show the remaining condition, if we choose any prime ideal \(\mathfrak{p}\) of \(A\), from

\[\mathfrak{m}_1\cdots\mathfrak{m}_k=0\subseteq \mathfrak{p}\]

we know that \(\mathfrak{m}_l\subseteq \mathfrak{p}\) must hold for some \(l\), and therefore \(\mathfrak{m}_l=\mathfrak{p}\). From this, all parts of the first condition hold.

From this, one can show that any Artinian ring is a finite product of local Artinian rings. For Noetherian rings, the following holds.

Theorem 5 For a Noetherian ring \(A\), \(A\) is a finite product of domains if and only if for every maximal ideal of \(A\), \(A_\mathfrak{m}\) is a domain.

Proof

First, let \(A\) be a finite product of domains, \(A=\prod A_i\). Then the units of \(A_i\), \(e_i\), satisfy \(\sum_i e_i=1\) and, for \(i\neq j\), \(e_ie_j=0\), so for any prime ideal of \(A\), \(\mathfrak{p}\), we have \(e_i\not\in \mathfrak{p}\) for a unique \(i\), and since this element \(e_i\), for \(i\neq j\), annihilates \(A_j\), \(A_\mathfrak{p}=(A_i)_\mathfrak{p}\) holds. (§Localization, ⁋Proposition 5)

Conversely, assume that the localization of \(A\) at any maximal ideal \(\mathfrak{m}\) is a domain, and let \(\{\mathfrak{q}_i\}\) be the minimal prime ideals of \(A\). Then by §Associated Primes, ⁋Theorem 7, which will be proved in the next post, \(\{\mathfrak{q}_i\}\) is a finite set, and accepting this, a natural map from \(A\) to the direct product of finitely many domains

\[A \rightarrow \prod_{i\in I} A/\mathfrak{q}_i\]

exists. Now let us show that this map is an isomorphism. By §Properties of Localization, ⁋Proposition 4, it suffices to localize at an arbitrary maximal ideal and show that the following map

\[A_\mathfrak{m} \rightarrow \left(\prod_{i\in I} A/\mathfrak{q}_i\right)_\mathfrak{m}\]

is an isomorphism. Meanwhile, by §Localization, ⁋Proposition 8, there is a one-to-one correspondence between the minimal prime ideals of \(A\) contained in \(\mathfrak{m}\) and the minimal prime ideals of \(A_\mathfrak{m}\). However, by assumption \(A_\mathfrak{m}\) is a domain, so \(A_\mathfrak{m}\) has a unique minimal prime ideal \((0)\), and therefore only one of the \(\mathfrak{q}_i\) is contained in \(\mathfrak{m}\). Now from this, since

\[\left(\prod_{i\in I} A/\mathfrak{q}_i\right)_\mathfrak{m}=(A/\mathfrak{q}_i)_\mathfrak{m}=A_\mathfrak{m}\]

the above map is an isomorphism.

Corollary 6 For a Noetherian ring \(A\) and a finitely generated \(A\)-module \(M\), the following are all equivalent.

1. \(M\) has finite length.
2. A product of suitable maximal ideals \(\prod_{i=1}^n \mathfrak{m}_i\) annihilates \(M\).
3. The prime ideals containing \(\ann(M)\) are maximal.
4. \(A/\ann(M)\) is Artinian.

Proof

First, assume the first condition and choose a composition series of \(M\). For each \(k\), since \(M_k/M_{k+1}\cong A/\mathfrak{m}_{k+1}\), we have \(\mathfrak{m}_{k+1}M_k\subseteq M_{k+1}\), and therefore the product of maximal ideals \(\mathfrak{m}_n\cdots\mathfrak{m}_1\) annihilates \(M\), which gives the second condition. Now assume the second condition. Then for any prime ideal containing \(\ann(M)\), \(\mathfrak{p}\), since \(\mathfrak{p}\supseteq\prod \mathfrak{m}_i\), for some \(i\) we have \(\mathfrak{p}\supseteq \mathfrak{m}_i\), and therefore \(\mathfrak{p}=\mathfrak{m}_i\). That the third condition implies the fourth condition is the equivalence of the first and third conditions of Theorem 4, and finally, assuming the fourth condition, \(A/\ann(M)\), by the second condition of Theorem 4, has finite length as an \(A/\ann(M)\)-module, and since \(M\) is a finitely generated \(A/\ann(M)\)-module, we obtain the desired result.

Then the following holds.

Corollary 7 Fix a Noetherian ring \(A\), a finitely generated \(A\)-module \(M\), and, containing \(\ann(M)\), a prime ideal \(\mathfrak{p}\). Then, as an \(A_\mathfrak{p}\)-module, \(M_\mathfrak{p}\) has finite length if and only if \(\mathfrak{p}\) is minimal among the primes containing \(\ann(M)\).

Proof

If \(\mathfrak{p}\) is a minimal prime ideal containing \(\ann(M)\), then the \(A_\mathfrak{p}\)-module \(M_\mathfrak{p}\) has finite length. This is because, considering the localization \(A_\mathfrak{p}\), the prime ideals of \(A_\mathfrak{p}\) are, by §Localization, ⁋Proposition 8, the prime ideals contained in \(\mathfrak{p}\), and then by the minimality of \(\mathfrak{p}\), the only prime ideal containing \(\ann(M)A_\mathfrak{p}\) is \(\mathfrak{p}A_\mathfrak{p}\), which is the (unique) maximal ideal of the local ring \(A_\mathfrak{p}\).

Conversely, suppose that \(M_\mathfrak{p}\) is an \(A_\mathfrak{p}\)-module of finite length. Then by Corollary 6, the prime ideals containing the annihilator of \(M_\mathfrak{p}\), \(\ann(M)A_\mathfrak{p}\), are all maximal, and since these in turn are in one-to-one correspondence via §Localization, ⁋Proposition 8 with the prime ideals containing \(\ann(M)\) and contained in \(\mathfrak{p}\), one can reverse the above argument.

In particular, for any ideal of \(A\), \(\mathfrak{a}\), considering the \(A\)-module \(A/\mathfrak{a}\), we have \(\ann(A/\mathfrak{a})=\mathfrak{a}\), and from this we obtain the following.

Corollary 8 For a Noetherian ring \(A\), an arbitrary ideal \(\mathfrak{a}\), and, containing \(\mathfrak{a}\), a prime ideal \(\mathfrak{p}\), the following are all equivalent.

1. \(\mathfrak{p}\) is minimal among the prime ideals containing \(\mathfrak{a}\).
2. \(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p}\) is Artinian.
3. In the localization \(A_\mathfrak{p}\), for all sufficiently large \(n\), \((\mathfrak{p}A_\mathfrak{p})^n\subseteq \mathfrak{a}A_\mathfrak{p}\) always holds.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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